Chapter 4 The Divergence Theorem - CUHK Mathematics

Chapter 4

The Divergence Theorem

In this chapter we discuss formulas that connects different integrals. They are

(a) Green’s theorem that relates the line integral of a vector field along a plane curve toa certain double integral in the region it encloses.

(b) Stokes’ theorem that relates the line integral of a vector field along a space curve toa certain surface integral which is bounded by this curve.

(c) Gauss’ theorem that relates the surface integral of a closed surface in space to a tripleintegral over the region enclosed by this surface.

All these formulas can be unified into a single one called the divergence theorem in termsof differential forms.

4.1 Green’s Theorem

Recall that the fundamental theorem of calculus states that

ˆ b

a

f ′(x) dx = f(b)− f(a) .

It relates the integral of the derivative of a function over an interval [a, b] to the endpointvalues of the function. In higher dimension we replace the function by a vector field.A possible two dimensional extension would be a formula relating the double integral ofsome quantity involving the partial derivatives of a vector field to the line integral of thevector field along its boundary curve. This is the content of Green’s theorem.

1

2 CHAPTER 4. THE DIVERGENCE THEOREM

Theorem 4.1. (Green’s Theorem) Let F = Mi + Nj be a C1-vector field in an openset G in the plane. Suppose C is a simple, closed curve in G and the set D it bounds liescompletely inside G. Then

¨D

(∂N

∂x− ∂M

∂y

)dA =

˛C

M dx+N dy , (4.1)

where C is oriented in the anticlockwise way.

A simple, closed curve divides the plane into two regions, one bounded and the otherunbounded. Here C bounds D means D is the bounded region.

Recall that line integral˛C

M dx+N dy =

˛C

F · dr ,

is called the circulation of F around the closed curve C. When F is the velocity vectorfield of some fluid, its circulation around a curve measures the amount of the fluid flowingaround the curve in unit time. When an admissible parametrization r : [a, b] 7→ C ischosen, the line integral can be evaluated by the formula

˛C

M dx+N dy =

ˆ b

a

F(r(t)) · r′(t) dt .

Proof. We will prove Green’s theorem in a special case, namely, D can be expressedsimultaneously in the following two ways:

D = (x, y) : f1(x) ≤ y ≤ f2(x), x ∈ [a, b]

andD = (x, y) : g1(y) ≤ x ≤ g2(y), y ∈ [c, d] .

Typical examples of such regions include ellipses and rectangles.

We shall show that ¨D

∂M

∂ydA = −

˛C

M dx , (4.2)

and ¨D

∂N

∂xdA =

˛C

N dy . (4.3)

Green’s theorem follows by adding (4.2) and (4.3) together.

The boundary curve C of D consists of the four curves:

C1 : r1(x) = (x, f1(x)), x ∈ [a, b] ,

C2 : r2(x) = (x, f2(x)), x ∈ [a, b] ,

4.1. GREEN’S THEOREM 3

γ2 : γ2(y) = (b, y), y ∈ [f1(b), f2(b)] ,

γ1 : γ1(y) = (a, y), y ∈ [f1(a), f2(b)] ,

where γ1 and γ2 may degenerate into points. We have C = C1 + γ2 − C2 − γ1.

By Fubini’s theorem

¨D

∂M

∂ydA =

ˆ b

a

ˆ f2(x)

f1(x)

∂M

∂y(x, y) dydx

=

ˆ b

a

M(x, y)∣∣∣f2(x)

f1(x)dx

=

ˆ b

a

M(x, f2(x))−M(x, f1(x)) dx

= −ˆC1−C2

M dx .

On the other hand, γ′1(y) = (0, 1), so

ˆγ1

M dx =

ˆ f2(a)

f1(a)

M(a, y)x′(y) dy = 0,

as x′(y) ≡ 0. By the same reasoning

ˆγ2

M dx = 0 ,

too. Therefore,

¨D

∂M

∂ydA = −

ˆC1−C2

M dx =

˛C1+γ2−C2−γ1

M dx = −˛C

M dx ,

and (4.1) follows. Similarly, we can prove (4.2).

When the region D is of more complicated geometry, one can use horizontal andvertical lines to decompose it into the union of regions of the above types. We will not gointo the details.

We will discuss four applications of Green’s theorem:

• Evaluation of line integrals,

• Study independence of path,

• An area formula,

• Localizing divergence and rotation.


The first application is illustrated in the following example.

Example 4.1. Evaluate ˛C

−y2 dx+ xy dy ,

where C is the boundary of the square at (0, 0), (1, 0), (1, 1) and (0, 1) in anticlockwisedirection.

A direct evaluation is not difficult, but tedious as it involves evaluating four lineintegrals. Instead we take advantage of Green’s theorem

˛C

−y2 dx+ xy dy =

¨R

(∂xy

∂x− ∂ − y2

∂y

)=

¨R

3y dA(x, y)

=

ˆ 1

0

ˆ 1

0

3y dxdy

=3

2.

Next, we return to the discussion on independence of path of vector fields in Chapter3. We established Theorem 3.4 which asserts that a vector field in Rn is conservative ifand only if the compatibility condition (3.9) holds (when n = 2). Now, by using Green’stheorem, we will present a more general result.

A region in Rn is called simply connected if it is connected and every closed curve lyingin it can be deformed continuously to a point inside the set itself. The entire plane, adisk, a convex set and more general a star-shaped region are examples of simply connectedsets in the plane. On the other hand, a punctured disk (a disk with the center removed)and an annulus are examples of connected but not simply-connected regions. Roughlyspeaking, simply connected regions are those connected regions which do enclose any holes.

Theorem 4.2. Let F = Mi +Nj be a C1-vector field in a simply connected region G inthe plane. It is conservative if and only if the compatibility condition holds:

∂M

∂y=∂N

∂x. (4.4)

This generalizes Theorem 3.4 where it is required the vector field to be defined in theentire space.

Proof. In Chapter 3, it was shown that (4.4) (that is (3.9)) holds when the vector fieldF is conservative. Conversely, under (4.4) a potential function were constructed under


the assumption that the line integrals along all simple curves connecting two points havethe same value. It suffices to verify this property in the present situation. Let γ1 andγ2 be two simple curves connecting point A to point B. When these two curves do notintersect, γ ≡ γ1 − γ2 forms a simple closed curve. Green’s theorem implies that

˛γ

M dx+N dy = 0 ,

hence ˆγ1

M dx+N dy =

ˆγ2

M dx+N, dy .

When γ1 and γ2 intersect, we may add another curve γ3 connecting A and B so thatit does not intersect γ1 and γ2. Thus γ1 − γ3 and γ2 − γ3 form simple closed curvesrespectively. Using

ˆγ1

M dx+N dy =

ˆγ3

M dx+N dy =

ˆγ2

M dx+N dy ,

we draw the same conclusion. Using this property one can define the potential of F as inthe proof of Theorem 3.4. The existence of a potential Φ shows that

ˆ B

A

F · t ds = Φ(B)− Φ(A) ,

along any path from A to B in G no matter the path is simple or not. We conclude thatF is conservative.

Green’s theorem is a formula relating the line integral of a curve to a double integralof the region it encloses. This observation leads to a formula expressing the area of theregion in terms of a boundary integral.

Let A be the area of the region enclosed by a simple closed curve C in the plane.Applying Green’s theorem to the vector field yi yields

A = −˛C

y dx . (4.5)

Similarly, choosing the vector field to be xj yields

A =

˛C

x dy . (4.6)

These two formulas together implies a more symmetric formula for the area:

A =1

2

˛C

−y dx+ x dy . (4.7)

Example 4.2. Find the area of one leaf of the four-leaf rose r = 3 sin 2θ.


The leaf in the first quadrant is ranged over θ ∈ [0, π/2]. In general, when a curveis parametrized by r(θ) in polar coordinates, it is r(θ) cos θi + r(θ) sin θj in cartesiancoordinates. We have

−y dx+ x dy

= (−r(θ) sin θ)(r′(θ) cos θ − r(θ) sin θ) + (r(θ) cos θ)(r′(θ) sin θ + r(θ) cos θ)dθ

= r2(θ) dθ .

Using the area formula, the area of one leaf is

1

2

ˆ π/2

0

9 sin2 2θ dθ =9π

8.

Example 4.3. Find the area of the loop in the folium of Decartes given by the zeros ofthe equation x3 + y3 = 3xy.

We first introduce a parametrization of the curve. Letting y = tx, we get

x =3t

1 + t2, y =

3t2

1 + t3.

One can verifies that the loop is described by t ∈ [0,∞). As t runs from 0 to ∞, the loopruns in the positive direction. You may look up Wiki for more information concerningthis curve.

We have

−y dx =−3t2

1 + t33[(1 + t3)− 3t3]

(1 + t3)2=−9t2(1− 2t2)

(1 + t3)3.

Therefore, the area of the loop is given by˛C

−y dx =

ˆ ∞0

−9t2(1− 2t2)

(1 + t3)3dt

= −3

ˆ ∞0

1− 2z

(1 + z)3dz

=3

2.

These formulas express the area enclosed by a curve in terms of the curve. It has inter-esting geometric consequence. For instance, together with Fourier series, the last formulacan be used to prove the classical isoperimetric inequality, that is, among all regions en-closed by a simple closed curve with the same perimeter, only the disk has the largest area.

Finally, recall that in Chapter 3 we introduce the concept of the circulation and theflux of a vector along a curve. Let F = M i +N j be C1-vector field defined on the simple


closed oriented curve C with the chosen tangent t and normal n. The circulation and theflux of F around C is defined to ˛

C

M dx+N dy ,

and ˛C

M dy −N dx ,

respectively. Green’s theorem suggests a way to define the circulation and the flux of avector field at a point. In other words, we can localize circulation and flux.

Let p be a point in some open set G ⊂ R2 where a C1-vector field F = P i + Qjis defined. Let C be a simple, closed curve anitclockwisely oriented enclosing p in itsinterior, and D the region it bounds. The quantity

1

|D|

˛C

M dx+N dy =1

|D|

¨D

(∂N

∂x− ∂M

∂y

)dA

→ ∂N

∂x(p)− ∂M

∂y(p) .

In view of this, we define the curl (or the rotation) of F at p to be

rot F(p) =

(∂N

∂x− ∂Q

∂y

)(p) .

Similarly, the flux of F across C is equal to

˛C

M dy −N dx .

By Green’s theorem,

1

|D|

˛C

M dy −N dx =1

|D|

¨D

(∂M

∂x+∂N

∂y

)dA

→ ∂M

∂x(p) +

∂N

∂y(p) .

Hence, we define the divergence (or flux density) of F at p to be

div F(p) =

(∂M

∂x+∂N

∂y

)(p) .

Example 4.4. Evaluate ˛C

y

x2 + y2dx+

−xx2 + y2

dy ,


where C is the ellipse x2/4 + y2/9 = 1 oriented in positive direction.

By a direct computation, the vector field

y

x2 + y2i +

−xx2 + y2

j

satisfies My = Nx. See the end of Section 3.6 in Chapter 3. However, since it is notdefined at the origin, we cannot appeal to Green’s theorem to conclude that this lineintegral vanishes. What we could do is to change it to an easier line integral. In fact,let Cr be a small circle centered at the origin and is contained inside C. We orient Crin clockwise direction and connect Cr to C by the line segment L which runs from (r, 0)to (2, 0). Then Γ = C − L + Cr + L forms a closed curve enclosing a simply-connecteddomain. Γ is not simple, but we can lift ±L up a little bit to make it simple. ApplyingGreen’s Theorem to this simple, closed curve and then passing to limit, we have

0 =

˛Γ

M dx+N dy

=

(ˆC

−ˆL

+

ˆCr

+

ˆL

)M dx+N dy

=

(ˆC

+

ˆCr

)M dx+N dy .

Therefore,

ˆC

y

x2 + y2dx+

−xx2 + y2

dy = −ˆCr

y

x2 + y2dx+

−xx2 + y2

dy

=

ˆ 2π

0

(cos θ cos θ + (− sin θ)(− sin θ)) dt

= 2π .

The trick of adding an artificial line segment to form a simply conneced region in thisexample leads us to the general form of Green’s Theorem. Let D be a region boundedby several simple, closed curves C1, C2, · · · , Cn where C1 is the outer one and the rest aresitting inside C1.

Theorem 4.3. Let F = P i +Qj be a C1-vector field in D. Then

¨D

(∂Q

∂x− ∂P

∂y

)dA =

n∑j=1

˛Cj

P dx+Qdy ,

where C1 is oriented in anticlockwise way and Cj, j ≥ 2, are in clockwise way.

4.2. STOKES’ THEOREM 9

4.2 Stokes’ Theorem

Stokes’ theorem is a curved version of Green’s theorem, where the two dimensional regionis replaced by a surface in space. Let F = M i + N j + Pk be a C1-vector field in someopen set G in space. The curl of the vector field F is given by

curl F =

(∂P

∂y− ∂N

∂z

)i +

(−∂P∂x

+∂M

∂z

)j +

(∂N

∂x− ∂M

∂y

)k .

As F is C1, the curl of F is a continuous vector field in G. When the vector field is aplanar one, that is, M = M(x, y), N = (x, y) and P ≡ 0, the curl of F becomes(

∂N

∂x− ∂M

∂y

)k ,

the term appearing in Green’s theorem.

We often use the notation ∇× F to denote the curl of F as suggested by the formalexpression by determinant ∣∣∣∣∣∣

i j k∂x ∂y ∂zM N P

∣∣∣∣∣∣ .Let S be an oriented surface in space whose boundary is a curve C. An orientation

assigned to C is based on the right hand rule: Let n be the unit normal of S and b bethe unit vector at C which is tangential and pointing inward to S and perpendicular toC. Then the unit tangent of C t is chosen such that t × b points to the same directionof n. When a person is walking along the boundary in the positive direction (t-direction)with his/her head pointing upward n, the surface should be lying on his left hand side.

Theorem 4.4. (Stokes’ Theorem) Let S be an oriented surface in space whose bound-ary is a simple, closed curve C. For a C1-vector field F on S,

¨S

∇× F · n dσ =

˛C

F · dr ,

where the orientation of C is described as above.

Proof. In the following it is more convenient to use (x, y, z) and (M,N,P ) instead ofxi + yj + zk and M i +N j + Pk.

Let r : D → S be a parametrization of S so that ru×rv points to the normal directionof S. It can be shown that the anticlockwise direction of the boundary curve C1 of D ismapped to C with the same orientation of C. Let

γ(t) = (u(t), v(t)) t ∈ [a, b] ,


be a parametrization for C1. Letting r(u, v) = (x(u, v), y(u, v), z(u, v)), the map

t 7→ (x(u(t), v(t)), y(u(t), v(t)), z(u(t), v(t))) , t ∈ [a, b] ,

gives a parametrization of the curve C. Using

dx = (xuu′ + xvv

′) dt, dy = (yuu′ + yvv

′) dt, dz = (zuu′ + zvv

′) dt ,

the left hand side of the Stokes’ theorem is˛C

F · dr

=

ˆ b

a

M(xuu′ + xvv

′) +N(yuu′ + yvv

′) + P (zuu′ + zvv

′) dt

=

ˆ b

a

[(Mxu +Nyu + Pzu)u′ + (Mxv +Nyv + Pzv)v

′] dt

≡ˆ b

a

(Au′ +Bv′) dt ,

where

A(u, v)

= M(x(u, v), y(u, v), z(u, v))xu(u, v) +N(x(u, v), y(u, v), z(u, v))yu(u, v) + P (x(u, v), y(u, v), z(u, v))zu(u, v) ,

and

B(u, v)

= M(x(u, v), y(u, v), z(u, v))xv(u, v) +N(x(u, v), y(u, v), z(u, v))yv(u, v) + P (x(u, v), y(u, v), z(u, v))zv(u, v) .

Appealing to Green’s theorem, the left hand side of the Stokes’ formula is equal to

ˆ b

a

(Au′ +Bv′) dt

=

˛C1

(A,B) · dγ

=

¨D

(Bu − Av) dA(u, v)

=

¨D

[(Mxxu +Myyu +Mzzu)xv +Mxvu + (Nxxu +Nyyu +Nzzu)yv +Nyvu

+(Pxxu + Pyyu + Pzzu)zv + Pzvu − (Mxxv +Myyv +Mzzv)xu −Mxuv

−(Nxxv +Nyyv +Nzzv)yu −Nyuv − (Pxxv + Pyyv + Pzzv)zu − Pzvu] dA(u, v)

=

¨D

[(Nx −My)(xuyv − xvyu) + (Px −Mz)(xuzv − xvzu) + (Py −Nz)(yuzv − yvzu)] dA .


On the other hand, (u, v) 7→ r = (x(u, v), y(u, v), z(u, v)) parametrizes S by D. Wehave

ru × rv = (yuzv − yvzu,−xuzv + xvzu, xuyv − xvyu) .

The right hand side of the formula is

¨S

∇× F · n dσ

=

¨D

(Py −Nz,−Px +Mz, Nx −My) · ru × rv dA(u, v)

=

¨D

[(Py −Nz)(yuzv − yvzu) + (−Px +Mz)(−xuzv + xvzu) + (Nx −My)(xuyv − xvyu)] dA(u, v) .

By comparing the left and the right hand sides of Stokes’ formula, we conclude that thetheorem holds.

Example 4.5. Evaluate the line integral

˛C

(y + z) dx+ (z + x) dy + x dz,

where C is the intersection of the cylinder x2 + y2 = 1 with the plane x + y + 2z = 10oriented in the anticlockwise direction.

The direct way of evaluation is to write down a parametrization for the curve C.Indeed, the projection of this curve onto the xy-plane is the circle x2 + y2 = 1. Hence anadmissible parametrization is

θ 7→ (cos θ, sin θ, (10− cos θ − sin θ)/2) , θ ∈ [0, 2π] .

But, here we would like to apply Stokes’ theorem. Observing that C is the boundaryof the oriented surface given by the graph of z = (10 − x − y)/2 over the unit diskD1 : x2 + y2 ≤ 1 with normal pointing upward. For F = (y+ z)i+ (z+x)j+xk, we have∇× F = −i. On the other hand, the upward normal is

n =−ϕxi− ϕyj + k√

1 + ϕ2x + ϕ2

y

=

√6

6(i + j + 2k) .


By Stokes’ theorem,

˛C

(y + z) dx+ (z + x) dy + x dz

=

¨S

−i · n dσ

=−√

6

6

¨S

dσ

= −¨D1

1

2dA ,

(dσ = (1 + |∇ϕ|)2dA =

√3

2dA

)= −π

2.

A very useful consequence of Stokes’ theorem is

Corollary 4.5. Suppose that a simple closed curve C in space bounds two surfaces S1

and S2 so that S = S1

⋃S2 forms a closed surface whose unit outward normal coincides

with the normals of S1 and S2. For any C1-vector field defined in an open set containingS, ¨

S2

∇× F · n dσ = −¨S1

∇× F · n dσ .

Proof. Let C1 and C2 be the respective oriented boundary curves for S1 and S2. It is clearthat C2 = −C2. By Stokes’ theorem,

¨S2

∇× F · n dσ =

˛C2

F · dr

=

˛−C1

F · dr

= −˛C1

F · dr

= −¨S1

∇× F · n dσ .

Example 4.6. Let Σ be the upper hemisphere x2 + y2 + z2 = 1, z ≥ 0, with normalpointing upward. Evaluate the integral

¨Σ

∇× F · n dσ ,

where F = yi + x2zj + exzk .


First of all, we calculate

∇× F = −x2i− zexzj + (2xz − 1)k .

Let S be the unit disk in the xy-plane. We regard it as a flat surface in space. Withnormal pointing down, Σ and S are two oriented surfaces sharing the same boundary. Bythe above corollary, ¨

Σ

∇× F · n dσ

= −¨S

∇× (yi + x2zj + exzk) · −k dσ

=

¨S

(2xz − 1) dσ .

An obvious parametrization of S is (x, y) 7→ (x, y, 0). Hence we continue to get¨Σ

∇× (yi + x2zj + exzk) · n dσ =

¨D1

− dA = −π.

Alternatively, we may use Stokes’ theorem to replace it by a line integral. Indeed, theboundary curve is simply the unit circle in the xy-plane which admits the parametrizationC : θ 7→ (cos θ, sin θ), θ ∈ [0, 2π] in positive direction. By Stokes’ theorem,¨

Σ

∇× F · n dσ =

˛C

F · dr

=

ˆ 2π

0

(sin θi + k) · (− sin θi + cos θj) dθ

= −ˆ 2π

0

sin2 θ dθ

= −π .

Next, we examine the meaning of the curl vector ∇ × F. Let p be a point in spaceand introduce a coordinates system using this point as the origin. Let ξ be a unit vectorand P be the plane perpendicular to it. Consider a simple closed curve C on P whichsurrounds the origin. Letting its enclosed region be D, by Stokes’ theorem,

1

|D|

˛C

M dx+N dy + P dz

1

|D

¨D

∇× F · ξ dσ

→ ∇× F(p) · ξ ,

as C shrinks to p. Hence the term ∇× F(p) · ξ measures the strength of rotation of thevector field F along the ξ-direction. This gives a meaning to the curl vector. From

∇× F · ξ = |∇ × F||ξ| cos θ ,


where θ ∈ [0, π] is the angle between the curl vector and ξ, we see that the strength ofrotation is maximal when ξ is equal to ∇× F or −∇× F.

Vector fields that are gradients of functions constitute a large class of irrotational vec-tor fields.

Theorem 4.6. Let Φ be C2-function in some open set G in space. Then F = ∇Φ isirrotational.

Proof. When F is the gradient of Φ, M = Φx, N = Φy, P = Φz. Then

My −Nx = Φxy − Φyx = 0 .

Similarly all other equations in the compatibility conditions hold.

In Chapter 3, it was shown that a conservative C1-vector field F = M i + N j + Pkmust satisfy the compatibility condition

∂N

∂x=∂P

∂z,∂M

∂z=∂P

∂x,∂M

∂y=∂N

∂x.

In other words, a conservative vector field must be irrotational. Now we show that anyconservative vector field arises in this way when it lies in a simply connected region. As aconsequence of Green’s theorem, we have seen this when the vector field is in the plane.

Theorem 4.7. An irrotational C1-vector field F in some simply connected region G isconservative.

Proof. It suffices to show that the line integral of F along any simple closed curve in Gvanishes. Let C be a simple closed curve in the region G. Since G is simply connected, Ccan be deformed into a point inside G. The deformation itself forms an oriented surfaceS taking C as its boundary. By Stokes’ theorem and the compatibility condition,

˛C

F · dr =

¨S

∇× F · n dσ = 0 .

To conclude this section, we consider oriented surfaces which are bounded not by asingle but by finitely many nonintersecting simple, closed curves. Just like Theorem 4.3,using the trick of cutting-up one can establish the following general Stokes’ theorem.

4.3. DIVERGENCE THEOREM 15

Theorem 4.8. Let F be a C1-vector field in some open set G in space. Let S be aC1-surface in G which is bounded by finitely many nonintersecting simple, closed curvesC1, · · · , Cn. Then ¨

S

∇× F · n dσ =n∑j=1

˛Cj

F · dr .

4.3 Divergence Theorem

Recall that there are two different formulation for Green’s theorem. Let F = M i+N j bea C1-vector field in some open set G and C a simple closed curve whose enclosed regionD sitting in G. Then ¨

D

(∂N

∂x− ∂M

∂y

)dA =

˛C

F · t ds ,

where t is the unit tangent along the anticlockwise direction. Alternatively, we can expressthe formula in the following way:

¨D

(∂M

∂x+∂N

∂y

)dA =

˛C

F · n ds ,

where n is the outward unit normal of C.

Stokes’ theorem may be viewed as a generalised version of the tangent form of theGreen’s theorem where a plane region is replaced by a surface in space. The divergencetheorem we are going to discuss may be regarded as a generalised version of the normalform where now a plane region is replaced by a region in space.

Theorem 4.9. Let F = Mi +Nj + Pk be a C1-vector field in an open set G in R3. LetS be a closed C1-surface which is the boundary of a bounded region Ω ⊂ G. Then

˚Ω

∇ · F dV =

‹S

F · n dσ ,

where n is the unit outward normal at S.

The divergence of a vector field, ∇ · F is given by

∇ · F =∂M

∂x+∂N

∂y+∂P

∂z.

Proof. We only consider the special case that Ω can be expressed simultaneously in thefollowing three ways:

(x, y, z) : (x, y) ∈ D, f1(x, y) ≤ z ≤ f2(x, y),


(x, y, z) : (y, z) ∈ D, g1(y, z) ≤ x ≤ g2(y, z),

and(x, y, z) : (x, z) ∈ D, h1(x, z) ≤ y ≤ h2(x, z) ,

where D is a region in the respective plane. For instance, rectangular boxes and ellipsoidsadmit such representations.

We will show that in the first case,

˚Ω

∂P

∂zdV =

‹S

Pk · n dσ . (4.8)

A similar argument will establish

¨Ω

∂M

∂xdV =

‹S

M i · n dσ ,

and ¨Ω

∂N

∂ydV +

‹S

N j · n dσ .

The divergence theorem follows by adding up these three identities.

Let us now prove (4.9). We may further assume that S = S1

⋃S2 where

S1 = (x, y, f1(x, y)) : (x, y) ∈ D, D2 = (x, y, f2(x, y) : (x, y) ∈ D .

Moreover, these two surfaces meet a curve whose projection to the xy-plane is a curve Cbounding D. We write

‹S

Pk · n dσ =

¨S1

Pk · n dσ +

¨S2

Pk · n dσ.

As

n =−f2xi− f2yj + k√

1 + |∇f2|2

on S2,

¨S2

Pk · n dσ =

¨S2

P√1 + |∇f2|2

dσ =

¨D

P (x, y, f2(x, y)) dA(x, y).

On the other hand,

n =f1xi + f1yj− k√

1 + |∇f1|2

on S1 (noting that it points downward),

¨S2

Pk · n dσ = −¨D

P (x, y, f1(x, y)) dA(x, y) .


Therefore, ‹S

Pk · n dσ =

¨D

(P (x, y, f2(x, y))− P (x, y, f1(x, y))) dA(x, y) .

On the other hand, ˚Ω

∂P

∂zdV

=

˚D

ˆ f2(x,y)

f1(x,y)

∂P

∂zdzdA

=

¨D

(P (x, y, f2(x, y))− P (x, y, f1(x, y))) dA(x, y) ,

and the desired result follows.

Example 4.7. Verify the divergence theorem for the vector field F = xi + yj + zk andΩ is the ball B, x2 + y2 + z2 ≤ a2.

The divergence of F is

∇ · F =∂

∂xx+

∂

∂yy +

∂

∂zz = 3 .

Therefore, ˚B

∇ · F dV =

˚B

3 dV = 3× 4

3πa3 = 4πa3 .

On the other hand, it is clear that the outward unit normal at a boundary point (x, y, z)is given by

n =1

a(xi + yj + zk) .

We have ‹∂B

F · n dσ =

‹∂B

(xi + yj + zk) · 1

a(xi + yj + zk) dσ

=1

a× a2

‹∂B

dσ

= 4πa3 .

Example 4.8. Find the flux of F = xyi + yzj + xzk outward through the surface of thecube cut from the first octant by the planes x = 1, y = 1 and z = 1.

Let Ω be the solid cube and C its boundary. By the divergence theorem, the flux isequal to ¨

C

F · n dσ =

˚Ω

∇ · F dV .


It suffices to calculate the triple integral which is obviously easier. We have∇·F = x+y+z.Therefore, the flux is equal to

˚Ω

(x+ y + z) dV =

ˆ 1

0

ˆ 1

0

ˆ 1

0

(x+ y + z) dxdydz =3

2.

The divergence of a vector field may be regarded as a mapping which maps a vectorfield to a function. Let p be a point in some open set where a C1-vector field F is defined.Consider a simple closed surface S which contains p in its interior D. The outward fluxof F across S is given by ‹

S

F · n dσ ,

where n is the unit outward normal. By the divergence theorem,

1

|D|

‹S

F · n dσ =1

|D|

˚D

∇ · F dV

→ ∇ · F(p) ,

as S shrinks to p . Hence ∇ · F may be viewed as a measurement of the flux of F atthe point p. When the vector field is two dimensional, its divergence reduces to the onediscussed in the last chapter.

The possibility of localizing the divergence is of fundamental important in physics. Asan illustration let us derive the equation of continuity. Image the space is occupied withsome fluid which is described in terms of its density δ and velocity u. Both are functionsof (x, y, z, t). We fix a region Ω in space and consider the mass of the fluid inside thisregion. As the density changes in time, the mass

M(t) =

˚Ω

δ(x, y, z, t) dV ,

is also a function of time. The rate of change in mass is given by

d

dtM(t) =

˚Ω

∂

∂tδ(x, y, z, t) dV . (4.9)

On the other hand, the fluid that leaves or enters through a small piece ∆σ on the surfaceof Ω, S, in time ∆t is given by

u · n∆σ∆t ,

where n is the unit outward normal. Hence the mass leaving or entering the region through∆σ is equal to

δu · n∆σ∆t .


It follows that the mass that leaving/entering S in instant time is equal to

d

dtM(t) = lim

∆t→0

M(t+ ∆t)−M(t)

∆t

= − lim∆t→0

1

∆t

˚S

δu · n∆σ∆t dσ

= −‹S

δu · n dσ.

Comparing with (4.9), we have˚

Ω

dδ

dt(x, y, z, t) dV = −

‹S

δu · n dσ .

Now, by the divergence theorem,˚

Ω

(dδ

dt+∇ · δu

)dV = 0 .

Since here Ω could be any region, one conclude that

dδ

dt+∇ · δu = 0 , (4.10)

must hold. This equation, called the equation of continuity, relating the density to thevelocity is a form of conservation of mass. Together with other equations such as con-servation of energy, etc, one obtain a full system of partial differential equations for thedensity and the velocity of the fluid.

In the divergence theorem stated above, it is assumed that the boundary of the regionis a single closed surface. When the boundary consists of several closed surfaces, we havethe following general divergence theorem.

Theorem 4.10. Let Ω be a region in space whose boundary is composed of finitely manyclosed C1-surfaces S1, S2, · · · , Sn. Then for any C1-vector field F in Ω,

˚Ω

∇ · F dV =n∑j=1

‹Sj

F · n dσ ,

where n is the unit outward normal to Sj.

This theorem is the space version of the planar case Theorem 4.3. Its proof is similarby cutting up the region. Since it is inconvenient to draw pictures here, you are referredto the text for details.

A vector field is called divergence-free if F = 0. The following result provides us witha large class of divergence vector fields.


Theorem 4.11. The divergence of a C1-curl vector field always vanishes.

Proof. Let F be a curl, that is, F = ∇ ×H for some vector field H = M i + N j + Pk.Then

∇ · F = ∇ · ((Nz − Py)i + (−Mz + Px)j + (My −Nx)k)

= (Nz − Py)x + (−Mz + Px)y + (My −Nx)z

= 0 .

Recall that the curl of a gradient vector field always vanishes, that is, ∇ ×∇Φ = 0.Now, the divergence of a curl vector field always vanishes as well. In the former one asks,when a vector field F satisfies ∇× F = 0, is F a gradient? The answer is yes when theunderlying region is simply connected. Here one may ask a similar question. Namely,when ∇ · F = 0, is there another vector field A such that F = ∇×A? It turns out theanswer is yes when the region is star-shaped.

According to the divergence theorem, a divergence-free vector field F satisfies

‹S

F · n dσ = 0 ,

over any simple, closed surface S. In other words, its flux across any closed surface van-ishes. A special case is the following theorem of Gauss which appears in the study ofelectricity.

The electric field generated by a point charge locating at the origin is given by

E =q

4πε0

1

r3(xi + yj + zk) , r =

√x2 + y2 + z2 ,

where q is the charge and ε0 is some positive constant. The electric field is divergencefree:

∇ · E =q

4πε0

(y2 + z2 − 2x2

r5+x2 + z2 − 2y2

r5+y2 + x2 − 2z2

r5

)= 0 .

Theorem 4.12. The flux of the electric field across any closed C1-surface enclosing theorigin is the same and equal to q/ε0.

Proof. Let Ω be the interior of S. Since the electric field is not well-defined at the origin,we cannot apply the divergence directly to Ω. Instead we consider the region Ωr which is


bounded by S and the small sphere Σr : x2 + y2 + z2 = r2. Then the electric field is C1

in Ωr and is of divergence free. By the divergence theorem,

0 =

˚Ωr

∇ · E dV =

‹S

E · n dσ +

‹Σr

E · n dσ ,

and the flux across S is equal to

−‹

Σr

E · n dσ .

When regarded as the inner boundary of Ω, the unit outer normal n at Σr points towardthe origin and is given by −(xi + yj + zk)/r. So,

−‹

Σr

E · n dσ =q

4πε0

‹Σr

1

r3(xi + yj + zk) · 1

r(xi + yj + zk) dσ

=q

4πε0

1

r4

‹Σr

(x2 + y2 + z2) dσ

=q

ε0

.

Chapter 4 The Divergence Theorem - CUHK Mathematics

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Transcript of Chapter 4 The Divergence Theorem - CUHK Mathematics